JEE MAIN - Mathematics (2016 (Offline) - No. 5)
Let two fair six-faced dice $$A$$ and $$B$$ be thrown simultaneously. If $${E_1}$$ is the event that die $$A$$ shows up four, $${E_2}$$ is the event that die $$B$$ shows up two and $${E_3}$$ is the event that the sum of numbers on both dice is odd, then which of the following statements is $$NOT$$ true?
$${E_1}$$ and $${E_2}$$ are independent.
$${E_2}$$ and $${E_3}$$ are independent.
$${E_1}$$ and $${E_3}$$ are independent.
$${E_1},$$ $${E_2}$$ and $${E_3}$$ are independent.
Explanation
Total possible outcome with two six faced dice = 62 = 36
When dice A shows up 4, the possible cases are
E1 = { (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) } = 6 cases
$$\therefore$$ $$P\left( {{E_1}} \right) = {6 \over {36}} = {1 \over 6}$$
When dice B shows up 2, the possible cases are
E2 = { (1, 2) (2, 2) (3, 2) (4, 2) (5, 2) (6, 2) } = 6 cases
$$P\left( {{E_2}} \right) = {6 \over {36}} = {1 \over 6}$$
$${E_1} \cap {E_2}$$ = Common in both in E1 and E2 = { (4, 2) }
$$P\left( {{E_1} \cap {E_2}} \right) = {1 \over {36}}$$
And $$P\left( {{E_1}} \right)$$.$$P\left( {{E_2}} \right)$$ = $${1 \over 6}$$.$${1 \over 6}$$ = $${1 \over 36}$$
$$\therefore$$$$P\left( {{E_1} \cap {E_2}} \right)$$ = $$P\left( {{E_1}} \right)$$.$$P\left( {{E_2}} \right)$$
$$\therefore$$ E1 and E2 are independent.
$${E_3}$$ = [ (1, 2), (1, 4), (1, 6),
(2, 1), (2, 3), (2, 5),
(3, 2), (3, 4), (3, 6),
(4, 1), (4, 3), (4, 5),
(5, 2), (5, 4), (5, 6),
(6, 1), (6, 3), (6, 5) ] = 18 cases
$${E_1} \cap {E_3}$$ = { (4, 1) (4, 3) (4, 5) } = 3 cases
$$\therefore$$ $$P\left( {{E_1} \cap {E_3}} \right) = {3 \over {36}} = {1 \over {12}}$$ = $${1 \over 6} \times {1 \over 2}$$ = $$P\left( {{E_1}} \right)$$ $$ \times $$ $$P\left( {{E_3}} \right)$$
$$\therefore$$ E1 and E3 are independent.
$${E_2} \cap {E_3}$$ = { (1, 2) (3, 2) (5, 2) } = 3 cases
$$\therefore$$ $$P\left( {{E_2} \cap {E_3}} \right) = {3 \over {36}} = {1 \over {12}}$$ = $${1 \over 6} \times {1 \over 2}$$ = $$P\left( {{E_2}} \right)$$ $$ \times $$ $$P\left( {{E_3}} \right)$$
$$\therefore$$ E2 and E3 are independent.
$${E_1} \cap {E_2} \cap {E_3}$$ = 0
$$\therefore$$ $$P\left( {{E_1} \cap {E_2} \cap {E_3}} \right)$$ = 0
$$P\left( {{E_1}} \right) \times $$ $$P\left( {{E_2}} \right) \times $$$$P\left( {{E_3}} \right)$$ = $${1 \over 6}$$ $$ \times $$ $${1 \over 6}$$ $$ \times $$ $${1 \over 2}$$ = $${1 \over {72}}$$
$$\therefore$$ $${E_1},$$ $${E_2}$$ and $${E_3}$$ are not independent.
$$\therefore$$ Option (D) is correct.
When dice A shows up 4, the possible cases are
E1 = { (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) } = 6 cases
$$\therefore$$ $$P\left( {{E_1}} \right) = {6 \over {36}} = {1 \over 6}$$
When dice B shows up 2, the possible cases are
E2 = { (1, 2) (2, 2) (3, 2) (4, 2) (5, 2) (6, 2) } = 6 cases
$$P\left( {{E_2}} \right) = {6 \over {36}} = {1 \over 6}$$
$${E_1} \cap {E_2}$$ = Common in both in E1 and E2 = { (4, 2) }
$$P\left( {{E_1} \cap {E_2}} \right) = {1 \over {36}}$$
And $$P\left( {{E_1}} \right)$$.$$P\left( {{E_2}} \right)$$ = $${1 \over 6}$$.$${1 \over 6}$$ = $${1 \over 36}$$
$$\therefore$$$$P\left( {{E_1} \cap {E_2}} \right)$$ = $$P\left( {{E_1}} \right)$$.$$P\left( {{E_2}} \right)$$
$$\therefore$$ E1 and E2 are independent.
$${E_3}$$ = [ (1, 2), (1, 4), (1, 6),
(2, 1), (2, 3), (2, 5),
(3, 2), (3, 4), (3, 6),
(4, 1), (4, 3), (4, 5),
(5, 2), (5, 4), (5, 6),
(6, 1), (6, 3), (6, 5) ] = 18 cases
$${E_1} \cap {E_3}$$ = { (4, 1) (4, 3) (4, 5) } = 3 cases
$$\therefore$$ $$P\left( {{E_1} \cap {E_3}} \right) = {3 \over {36}} = {1 \over {12}}$$ = $${1 \over 6} \times {1 \over 2}$$ = $$P\left( {{E_1}} \right)$$ $$ \times $$ $$P\left( {{E_3}} \right)$$
$$\therefore$$ E1 and E3 are independent.
$${E_2} \cap {E_3}$$ = { (1, 2) (3, 2) (5, 2) } = 3 cases
$$\therefore$$ $$P\left( {{E_2} \cap {E_3}} \right) = {3 \over {36}} = {1 \over {12}}$$ = $${1 \over 6} \times {1 \over 2}$$ = $$P\left( {{E_2}} \right)$$ $$ \times $$ $$P\left( {{E_3}} \right)$$
$$\therefore$$ E2 and E3 are independent.
$${E_1} \cap {E_2} \cap {E_3}$$ = 0
$$\therefore$$ $$P\left( {{E_1} \cap {E_2} \cap {E_3}} \right)$$ = 0
$$P\left( {{E_1}} \right) \times $$ $$P\left( {{E_2}} \right) \times $$$$P\left( {{E_3}} \right)$$ = $${1 \over 6}$$ $$ \times $$ $${1 \over 6}$$ $$ \times $$ $${1 \over 2}$$ = $${1 \over {72}}$$
$$\therefore$$ $${E_1},$$ $${E_2}$$ and $${E_3}$$ are not independent.
$$\therefore$$ Option (D) is correct.
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